Question

What will be the value of formation of $C{O_2}$ .
$C{O_{\left( {\text{g}} \right)}} + \dfrac{1}{2}{{\text{O}}_{2(g)}} \to C{O_{2\left( {\text{g}} \right)}} + {\text{x}}\;KCal$ $$
${{\text{C}}_{\left( {\text{s}} \right)}} + {{\text{O}}_{2\left( {\text{g}} \right)}} \to C{O_{2\left( {\text{g}} \right)}} + 2{\text{y}}\;KCal$
(A) $2y + x$
(B) $2y - x$
(C) $- (x + 2y)$
(D) $x - 2y$

Answer 1

Hint : Here 2 reactions are given both forming Carbon Dioxide gas. In the first reaction one mole of carbon monoxide gas reacts with half mole of oxygen gas to give 1 mole of carbon dioxide gas releasing x $KCal$ of energy. In the second reaction one mole of solid carbon reacts with 1 mole of oxygen gas to give 1 mole of carbon dioxide gas releasing 2y $KCal$ energy. Both reactions are exothermic as energy is released.

Complete Step By Step Answer:
Now we will see how we can find the total heat of formation of carbon dioxide from given reactions
$C{O_{\left( {\text{g}} \right)}} + \dfrac{1}{2}{{\text{O}}_{2(g)}} \to C{O_{2\left( {\text{g}} \right)}} + {\text{x}}\;KCal$ ………. (1)
let’s take the above reaction as reaction (1) with enthalpy $\vartriangle {H_1}$ = ${\text{x}}\;KCal$
${{\text{C}}_{\left( {\text{s}} \right)}} + {{\text{O}}_{2\left( {\text{g}} \right)}} \to C{O_{2\left( {\text{g}} \right)}} + 2{\text{y}}\;KCal$ ……………. (2)
let’s take the above reaction as reaction (2) with enthalpy $\vartriangle {H_2}$ = $2{\text{y}}\;KCal$
Reversing the order of reaction (1) we get,
$C{O_{2\left( {\text{g}} \right)}} \to C{O_{\left( {\text{g}} \right)}} + \dfrac{1}{2}{{\text{O}}_{2(g)}}{\text{ - x}}\;KCal$ …………. (3)
let’s take the above reaction as reaction (3) with enthalpy $\vartriangle {H_3}$ = ${\text{ - x}}\;KCal$
Now, if we add reaction (2) and (3), we get
${C_{\left( s \right)}} + \dfrac{1}{2}{{\text{O}}_{2(g)}} \to C{O_{2\left( {\text{g}} \right)}}$ ……………. (4)
let’s assume the above reaction as reaction (4)
Now to calculate enthalpy of above reaction, $\vartriangle {H_4}$
$\vartriangle {H_4} = \vartriangle {H_2} + \vartriangle {H_3}$
Putting values, in the equation
$\vartriangle {H_4} = 2{\text{y}}\;KCal + ({\text{ - x}}\;KCal)$
we get,
$\vartriangle {H_4} = (2{\text{y}}\;{\text{ - x)}}\;KCal$
So, enthalpy of formation of $C{O_2}$ or Heat of formation of $C{O_2}$ is $2y - x$ .
Therefore option (B) is the correct answer.

Note :
Be very careful when you are using reaction to solve this type of problem, we try to solve our problem by reversing reaction this changes the enthalpy like in above reaction, by performing additions or subtractions of reaction this can be very confusing, to easily solve that you should try numbering your reactions and use them accordingly.